JEE Advanced 2012 Paper 1Question 1
Let $f(x)=\begin{cases} x^2\left|\cos\dfrac{\pi}{x}\right|, & x\neq 0\\ 0, & x=0\end{cases}$, $x\in\mathbb{R}$, then $f$ is
Differentiation and its Application
Solution
Concept. At $0$ use the definition; at $2$, $\cos\frac\pi x$ changes sign, so the modulus creates a corner. Formulas used. (1) $\dfrac{d}{dx}\Big(x^2\cos\dfrac\pi x\Big)=2x\cos\dfrac\pi x+\pi\sin\dfrac\pi x$Step 1 - At $x=0$. $$\left|\frac{f(x)-f(0)}{x}\right|=|x|\left|\cos\tfrac\pi x\right|\le|x|\to0.$$ So $f'(0)=0$: differentiable. Step 2 - At $x=2$. For $x$ slightly below $2$, $\frac\pi x>\frac\pi2$ and $\cos\frac\pi x<0$, so $f=-x^2\cos\frac\pi x$. For $x$ slightly above $2$, $f=x^2\cos\frac\pi x$. Step 3 - Slopes. By (1) the derivative of $x^2\cos\frac\pi x$ at $2$ is $0+\pi=\pi$. So the left slope is $-\pi$, the right slope is $\pi$. Not differentiable at $2$. Answer: (B)